how can I divide number (money) to x number equally the number could be with one or two decimal or without it
such as 1000 or 100.2 or 112.34
I want to be able to split that number into x part all of them equally, however if it's not odd number the extra number to the last one.
for example
3856 / 3
1285.33
1285.33
1285.34
review of other answers
After re-examining the solutions provided here, I noticed that they produce some strange results.
@GeorgeMauler's divideCurrencyEqually appears to work with the inputs provided in the original question. But if you change them up a bit, it produces a very strange result for …
// it appears to work
divideCurrencyEvenly(10,3)
=> [ '3.33', '3.33', '3.34' ]
// wups ...
divideCurrencyEvenly(10,6)
=> [ '1.67', '1.67', '1.67', '1.67', '3.32' ]
// hmm this one works ...
divideCurrencyEvenly(18,5)
=> [ '3.60', '3.60', '3.60', '3.60', '3.60' ]
// wups ...
divideCurrencyEvenly(100,7)
=> [ '14.29', '14.29', '14.29', '14.29', '14.29', '28.55' ]
While @Barmar's solution seems to perform a bit better … well a little bit …
// it appears to work
divide(10,3)
=> [ '3.33', '3.33', 3.34 ]
// acceptable, i guess
divide(10,6)
=> [ '1.66', '1.66', '1.66', '1.66', '1.66', 1.700000000000001 ]
// hmm, not so good, but still acceptable
divide(18,5)
=> [ '3.60', '3.60', '3.60', '3.60', 3.5999999999999996 ]
// as acceptable as the last two, i guess
divide(100,7)
=> [ '14.28', '14.28', '14.28', '14.28', '14.28', '14.28', 14.320000000000007 ]
Numbers like 3.5999999999999996 are mostly forgivable because that's how float arithmetic works in JavaScript.
But, what I find most disturbing about divide is that it's giving a mixed-type array (string and float). And because the numbers aren't rounded (to the correct precision) in the output, if you were to add the result up on paper, you will not arrive back at your original numerator input — that is, unless you do the rounding on your result.
and we're still fighting for equality …
My last grievance exists for both of the above solutions too. The result does not represent a distribution that is as close to equal as is possible.
If you divide 100 by 7 with a precision of 2, @Barmar's solution (with the rounding problem fixed) would give …
[ '14.28', '14.28', '14.28', '14.28', '14.28', '14.28', 14.32 ]
If these were people and the numbers represented monies, 1 person shouldn't pay 4 pennies more. Instead 4 people should pay 1 penny more …
[ 14.29, 14.29, 14.29, 14.29, 14.28, 14.28, 14.28 ]
That's as close to equal as is possible
back to square one
I was dissatisfied with the solutions, so I made one of my own, distribute.
It has 3 parameters: precision p, divisor d, and numerator n.
It scales the numerator and finds the largest integer q where q * d <= n. Using modular division we know how many "slices" need to contribute q+1. Lastly, each q or q+1 is scaled back down and populates the output array.
const fill = (n, x) =>
Array (n) .fill (x)
const concat = (xs, ys) =>
xs .concat (ys)
const quotrem = (n, d) =>
[ Math .floor (n / d)
, Math .floor (n % d)
]
const distribute = (p, d, n) =>
{ const e =
Math .pow (10, p)
const [ q, r ] =
quotrem (n * e, d)
return concat
( fill (r, (q + 1) / e)
, fill (d - r, q / e)
)
}
console .log
( distribute (2, 3, 10)
// [ 3.34, 3.33, 3.33 ]
, distribute (2, 6, 10)
// [ 1.67, 1.67, 1.67, 1.67, 1.66, 1.66 ]
, distribute (2, 5, 18)
// [ 3.6, 3.6, 3.6, 3.6, 3.6 ]
, distribute (2, 7, 100)
// [ 14.29, 14.29, 14.29, 14.29, 14.28, 14.28, 14.28 ]
)You'll see that I made precision a parameter, p, which means you can control how many decimal places come out. Also note how the largest difference Δ between any number in the result is Δ <= 1/10^p
distribute (0, 7, 100)
=> [ 15, 15, 14, 14, 14, 14, 14 ] // Δ = 1
distribute (1, 7, 100)
=> [ 14.3, 14.3, 14.3, 14.3, 14.3, 14.3, 14.2 ] // Δ = 0.1
distribute (2, 7, 100)
=> [ 14.29, 14.29, 14.29, 14.29, 14.28, 14.28, 14.28 ] // Δ = 0.01
distribute (3, 7, 100)
=> [ 14.286, 14.286, 14.286, 14.286, 14.286, 14.285, 14.285 ] // Δ = 0.001
distribute (4, 7, 100)
=> [ 14.2858, 14.2857, 14.2857, 14.2857, 14.2857, 14.2857, 14.2857 ] // Δ = 0.0001
distribute can be partially applied in meaningful ways. Here's one way petty people could use it to precisely split the bill at a restaurant …
// splitTheBill will use precision of 2 which works nice for monies
const splitTheBill = (people, money) =>
distribute (2, people, money)
// partyOfThree splits the bill between 3 people
const partyOfThree = money =>
splitTheBill (3, money)
// how much does each person pay ?
partyOfThree (67.89)
=> [ 18.93, 18.93, 18.92 ]
And here's an effective way to divide people into groups — while being careful not to divide an individual person — which typically results in death …
// p=0 will yield only whole numbers in the result
const makeTeams = (teams, people) =>
distribute (0, teams, people)
// make 4 teams from 18 people
// how many people on each team?
makeTeams (4, 18)
=> [ 5, 5, 4, 4 ]
function divide(numerator, divisor) {
var results = [];
var dividend = (Math.floor(numerator/divisor*100)/100).toFixed(2); // dividend with 2 decimal places
for (var i = 0; i < divisor-1; i++) {
results.push(dividend); // Put n-1 copies of dividend in results
}
results.push(numerator - (divisor-1)*dividend); // Add remainder to results
return results;
}
Sounds like a pretty straightforward loop/recursion.
function divideEvenly(numerator, minPartSize) {
if(numerator / minPartSize< 2) {
return [numerator];
}
return [minPartSize].concat(divideEvenly(numerator-minPartSize, minPartSize));
}
console.log(divideEvenly(1000, 333));
To get it to be two decimals of currency multiply both numbers by 100 before calling this function then divide each result by 100 and call toFixed(2).
function divideCurrencyEvenly(numerator, divisor) {
var minPartSize = +(numerator / divisor).toFixed(2)
return divideEvenly(numerator*100, minPartSize*100).map(function(v) {
return (v/100).toFixed(2);
});
}
console.log(divideCurrencyEvenly(3856, 3));
//=>["1285.33", "1285.33", "1285.34"]
Since we are talking about money, normally a few cents different doesn't matter as long as the total adds up correctly. So if you don't mind one payment possibly being a few cents greater than the rest, here is a really simple solution for installment payments.
function CalculateInstallments(total, installments) {
// Calculate Installment Payments
var pennies = total * 100;
var remainder = pennies % installments;
var otherPayments = (pennies - remainder) / installments;
var firstPayment = otherPayments + remainder;
for (var i = 0; i < installments; i++) {
if (i == 0) {
console.log("first payment = ", firstPayment / 100);
} else {
console.log("next payment = ", otherPayments / 100);
}
}
}Since you said you wanted the highest payment at the end you would want to modify the if statement in the for loop: if (i==installments-1) { //last payment }
There is an issue on @user633183's distribute but only happen when the divider is lower than 3.
distribute(2, 2, 560.3)
// [ '280.15' , '280.14']
distribute(2, 1, 74.10)
// [ '74.09' ]
distribute(2, 1, 74.60)
// [ '74.59' ]
I rewrote the answer by Guffa into javascript
const distribute = (precision, divider, numerator) => {
const arr = [];
while (divider > 0) {
let amount = Math.round((numerator / divider) * Math.pow(10, precision)) / Math.pow(10, precision);
arr.push(amount);
numerator -= amount;
divider--;
}
return arr.sort((a, b) => b-a);
};
Here are the results
distribute(0, 7, 100)
=> [ 15, 15, 14, 14, 14, 14, 14 ]
distribute(1, 7, 100)
=> [ 14.3, 14.3, 14.3, 14.3, 14.3, 14.3, 14.2 ]
distribute(2, 7, 100)
=> [ 14.29, 14.29, 14.29, 14.29, 14.28, 14.28, 14.28 ]
distribute(3, 7, 100)
=> [ 14.286, 14.286, 14.286, 14.286, 14.286, 14.285, 14.285 ]
distribute(4, 7, 100)
=> [ 14.2858, 14.2857, 14.2857, 14.2857, 14.2857, 14.2857, 14.2857 ]
// and of course
distribute(2, 2, 560.3)
=> [ 280.15, 280.15 ]
distribute(2, 1, 74.10)
=> [ 74.1 ]
distribute(2, 1, 74.60)
=> [ 74.6 ]
please check the below code,
function Dividently(Amount, Quantity) {
var m = Amount* 100,
n = m % Quantity,
v = Math.floor(m / Quantity) / 100,
w = Math.floor(m / Quantity+ 1) / 100;
for (var i = 0, out = new Array(Quantity); i < Quantity; ++i) {
out[i] = i < n ? w : v;
}
return out;
}
var arr = Dividently(3856, 3);
Math.floor() - Round a number downward to its nearest integer
If anyone is looking for a randomly distributed vanilla JS solution here is mine:
function fairDivision(resolution, numerator, denominator) {
// preserves numerator(n) integrity when dividing into denominator(d) bins
// remainders are randomly distributed into sequential bins
// resolution is number of significant digits after decimal
// (-'ve resolution for insignificant digits before decimal).
const n = numerator * Math.pow(10, resolution);
const remainder = n % denominator;
const base = (n - remainder) / denominator;
const offset = Math.floor(Math.random()*denominator); // index of first 'round-up' bin
let arr = []
let low= (offset + remainder) % denominator;
let a; let b; let c = (low < offset); //logical units
for (let i=0; i < denominator; i++) {
a = (i < low);
b = (i >= offset);
if ((a && b) || (a && c) || (b && c)) {
arr.push(base +1)
} else{
arr.push(base)
}
arr[i] = arr[i] / Math.pow(10, resolution);
}
return arr;
}
This is based in @Thank_you's conclusions, just to divide portions with no decimals, based in equality:
//n=how many groups, lot = total units
function makeTeams2(n,lot){
var div = lot/n;
var portion = Math.floor(div);
var remains = Math.round(n*((div) % 1));
var arr = [];
for(var i=0; i<n; i++) arr.push(portion);
for(var i=0; i<remains; i++) arr[i] = arr[i] + 1;
return arr;
}
// >> makeTeams2(7,100);
// >> [15,15,14,14,14,14,14]
This is a little faster than @Thank_you's "makeTeams(teams,people)" constant but I'm accepting performance improvements since I'm a beginner, I just did this on my own way and wanted to share.
<button onclick="myFunction(3856,3)">Try it</button>
function myFunction(number,divide) {
var money = number / divide;
money = Math.ceil(money * 100) / 100;
alert(money); //1285.34
}
